What is the probability that 2 or more numbers out of 4 are among the drawn lottery numbers (6 out of 49 without replacement)? I have the following combinatoric problem to solve; I have searched and couldn't find an answer already existing.
In the main lottery in Germany, 6 numbers are to be drawn out of 49 {1,2,…49} without replacement. Let's assume I can only choose 4 numbers between 1 and 49. What is the probability that at least 2 of the four numbers are among the six numbers that have been drawn in the lottery?
I drew a sample of numbers and therefore have a rough idea that it should be around 6.2-7% if I did everything correctly and the sample was large enough. However, I am interested in the exact probability and its derivation.
Any help would be highly appreciated.
 A: There are $6$ winning numbers and $43$ "non-winning" numbers.
To get only $2$ winning numbers out of the $4$ you choose, using the standard hypergeometric formula for choosing w/o replacement, there are $\binom62\binom{43}2$ ways out of a total of $\binom{49}4$ ways
Proceeding similarly, $\Bbb P(2,3\;or\;4\; winning) = \dfrac{\binom 6 2\binom{43}2 +\binom 6 3\binom{43}2 +\binom 6 4\binom{43}0}{\binom{49}4} = \dfrac{515}{7567} \approx 0.06806$ 
A: We can consider three cases:


*

*Draw four numbers out of six. All four balls must belong to the six chosen balls, so the probability of this happening equals:
$$\frac{6}{49} \cdot \frac{5}{48} \cdot \frac{4}{47} \cdot \frac{3}{46}$$

*Draw three numbers out of six. One way to do this, is by first choosing three balls belonging to the six chosen balls, followed by one ball which does not belong to these balls. This probability equals:
$$\frac{6}{49} \cdot \frac{5}{48} \cdot \frac{4}{47} \cdot \frac{43}{46}$$
Since there are four different turns in which we can choose a ball not belonging to the six chosen balls, the probability of selecting three balls out of six equals:
$${4 \choose 1} \cdot \frac{6}{49} \cdot \frac{5}{48} \cdot \frac{4}{47} \cdot \frac{43}{46}$$

*Draw two numbers out of six. This time there are ${4 \choose 2} = 6$ combinations of turns in which we can choose a ball not belonging to the six chosen balls, so this happens with probability:
$${4 \choose 2} \cdot \frac{6}{49} \cdot \frac{5}{48} \cdot \frac{43}{47} \cdot \frac{42}{46}$$
Adding this all up, the probability of choosing at least two correct numbers equals:
$$\frac{6 \cdot 5 \cdot 4 \cdot 3 + 4 \cdot 6 \cdot 5 \cdot 4 \cdot 43 + 6 \cdot 6 \cdot 5 \cdot 43 \cdot 42}{49 \cdot 48 \cdot 47 \cdot 46} \approx 0.0681$$
